Midpoint - comprehensive study approach

In grouped data, there need a value that can be used in each group so that we can get ∑fx for each class and eventually summation of values to get the mean. The value used to represent values of a given class is the midpoint (x) obtained by adding lower and upper class boundary and dividing them by two.

Example

The frequency table below shows masses in kilograms of some grade 10 students.

class	Frequency
60-64	8
65-69	7
70-74	12
75-79	9
80-84	6
85-89	5

a table showing frequency against group of ages

Required:

(a) State the modal class
(b) Estimate the mean
(c) Estimate the median

solution

(a) The modal class is 70-74 because it has the highest frequency (12).

(b) we redraw the table to include a column for the midpoint values

the midpoint of class 60-64=(60+64)/2 = 62, and you will follow the same step to get midpoints of other classes which will include 67,72,77,82 and 87.

Mean x̄ = (∑fx)/(∑f)

hence Mean x̄ = (3449)/(47) = 574.833

(c) The median value is the value at the 24th position of the frequency. This is because of all the data were arranged in ascending order, the middle one would be at the 24th position. On the left of 24, there would be 23 values and on the right there would be 23 values.

if we add the frequencies cumulatively we have 8, 15, 27, 36, 42, 47 respectively. It means there are 27 values with 74 kg and below, the middle value (24^th value) is found in the class 70-74.

The lower class boundary of the median class is 69.5 kg. which best fits the mass of the 15th person.

to get the mass of the 24th person ,we need 9 people from the median class.

so we get 69.5 + ((24-15)/12) * 5 = 69.5+3.75 = 73.25

Exercise

(a) The table below shows the masses of some people that visited a hospital

Class	Mid-point(x)	frequency f	fx
51-55		9
56-60		13
61-65		15
66-70		17
71-75		24

a table of masses of people that visited a hospital

(a) copy and complete the table

(b) Find:

The mean
The Median

(b) The height in centimeters of 25 people were measured as follows:

156, 170, 185, 167, 179, 180, 174, 169, 169, 162, 159, 162, 165, 179, 174, 175, 184, 189, 183, 190, 165, 156, 158, 169, 162.

(a) Using a class width of 5 make a grouped frequency table

(b) from the table estimate:

The mean
The median

solution

(a) The lowest value is 156 and so the first group will be 156-160, the highest value is 190. The range is 190-156=34.

34/5=6.8 ≈ 7.

So there is about 7 groups. We develop a table as shown

(b) The formula for calculating the mean is

∑fx=4265 as read from the table

∑f = 25

hence

The total number of items, which is the summation of frequency (∑f) is 25.

The median, which is the number at the center is the 13^thvalue

The class that will contain 13th value is a class that starts with 10th and ends with 14th value as shown in the table. That class is 166-170 and it is the median class.

The lower class boundary of the median class is 165.5

The cumulative frequency above the median class is 9

The frequency of the median class is 5

The formula for the median is given as

where L is the lower class boundary of the median class, n the total frequency, C the cumulative frequency below the median class, i the class width and f the frequency of the median class

Hence the median of the data is 169

29, 23, 22, 26, 35, 38, 45, 42, 30, 22, 35, 33, 34, 29, 36, 27, 38, 40, 44, 39, 38, 43, 40, 35, 30, 25, 29, 26, 28, 33, 34

Required:

(a) using a class width of 5, make a frequency table for the data

(b) Find the mean and the median for the data

Tag: Midpoint

Grouped Data

Example

solution

Exercise

solution

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